Integral of x*dx/(x+1) dx

1. Rewrite the integrand:

   $$x 1 \cdot \frac{1}{x + 1} = 1 - \frac{1}{x + 1}$$

2. Integrate term-by-term:

   1. The integral of a constant is the constant times the variable of integration:

      $$\int 1\, dx = x$$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \left(- \frac{1}{x + 1}\right)\, dx = - \int \frac{1}{x + 1}\, dx$$

      1. Let $u = x + 1$.

         Then let $du = dx$ and substitute $du$:

         $$\int \frac{1}{u}\, du$$

         1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

         Now substitute $u$ back in:

         $$\log{\left(x + 1 \right)}$$

      So, the result is: $- \log{\left(x + 1 \right)}$

   The result is: $x - \log{\left(x + 1 \right)}$

3. Add the constant of integration:

   $$x - \log{\left(x + 1 \right)}+ \mathrm{constant}$$

The answer is: